Q48 / 180 · NEET 2019 (Odisha) · Chemistry
An ideal gas expands isothermally from $10^{-3} \mathrm{~m}^3$ to $10^{-2} \mathrm{~m}^3$ at $300 \mathrm{~K}$ against a constant pressure of $10^5 \mathrm{Nm}^{-2}$. The work done on the gas is
- A$+270 \mathrm{~kJ}$
- Correct answerB$-900 \mathrm{~J}$
- C$+900 k J$
- D$-900 k J$
Solution
For an isothermal irreversible expansion, Work done $(\mathrm{W})=-\mathrm{p}_{\mathrm{ext}}\left(\mathrm{V}_2-\mathrm{V}_1\right)$ where, $\mathrm{V}_1=$ initial volume
$$
\mathrm{V}_2=\text { final volume }
$$
Given, $\mathrm{p}_{\mathrm{ext}}=10^5 \mathrm{Nm}^{-2}, \mathrm{~V}_1=10^{-3} \mathrm{~m}^3, \mathrm{~V}_2=10^{-2} \mathrm{~m}^3$
On substituting the given values in Eq, (i),
We get,
$$
\begin{aligned}
W & =-10^5 \mathrm{Nm}^{-2}\left(10^{-2} \mathrm{~m}^3-10^{-3} \mathrm{~m}^3\right) \\
& =-10^5 \mathrm{Nm}^{-2} \times 10^{-3}(10-1) \mathrm{m}^3 \\
& =-900 \mathrm{Nm}=-900 \mathrm{~J}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.